Solve the inequality for a, given x=2 is a solution

Question

Find all the values of a that make x=2 a solution to:

(3a+x)/(x-3a)<1

Answer 1

If x=2 is a solution: 

(3a+2)/(2-3a) < 1

Since we are dealing with inequalities, there are two cases to consider. These are: 

Case 1: 2-3a > 0

Case 2: 2a -3a < 0

The reason for the two is that when you multiply both sides of and inequality by a positive number the inequality is the same. When multiplied by a negative number the inequality is reversed.  

Case 1: 2-3a > 0

(3a+2)/(2-3a) < 1

3a + 2 < 2 – 3a

6a < 0

a < 0

check against  2 – 3a > 0 to make sure this is still true

2 > 3a

2/3 > a

a < 2/3

for this case, a < 0 and a < 2/3 must be true. So, a  < 0

Case 2: 2a -3a < 0

(3a+2)/(2-3a) < 1

3a + 2 > 2 -3a

6a > 0

a > 0

check against  2 – 3a < 0 to make sure this is still true

2 < 3a

2/3 < a

a > 2/3

for this case, a > 0 and a >2/3 must be true. So, a >2/3

This makes the solution set {a| a < 0 || a > 2/3}

Answer 2

(3a+x)/(x-3a)<1

(3a+2)/(2-3a)<1

3a+2 < 2-3a

6a< 0

a<0